Problem Solving Strategies

yes

And in your frame times passes at one second per second.

yes

Where things get weird is that the coordinate systems for different observers my not match up.

If all the observers are stationary then their coordinate systems differ only by the choice of origin e.g. if I'm a metre away from you my distances will differ from yours by one metre.

6:26 AM

suppose i keep an axis exclusively for time instead of cloak, so can we see time dilation as a change in angle between two time co-ordinates of two different observers..

Basically yes. If I am observing you and you are moving then I would find your time axis is rotated relative to mine.

And your space axis is also rotated relative to mine.

Looks like now I am getting a proper view on relativity

This is the origin of time dilation and length contraction.

@JohnRennie yes...

Have you heard of the Lorentz transformations?

6:31 AM

@JohnRennie I heard only about lorentz gamma

p = mv*gamma

gamma = 1/root(1-(v/c)^2) ryt ?

but i have no idea on transformation

Suppose I am using coordinates (t, x, y, z) and you are using coordinates (t', x', y', z') then the Lorentz transformations convert my coordinates to yours and vice versa.

@JohnRennie it tells the difference ?

Suppose a supernova goes off and you and I measure its position. I measure its position to be (t, x, y, z) and you measure its position to be (t', x', y' z').

ohh

Then the LTs describe how our measurements are related i.e. how my t compares to your t', my x compares to your x', and so on.

Time dilation happens when my t and your t' are different, so the LTs can be used to calculate time dilation.

6:37 AM

And this provokes me to ask a quesion.. does time dilation affects light ?

You need to be clear what you are asking. Firstly time dilation isn't some active "force" that affects things. It's just a result of different observers measuring the same thing in different ways. You probably already knew that, but it's worth making it absolutely clear.

Now are you asking how a light ray experiences time?

exactly

The answer is that there is no answer to that question because anything moving at the speed of light has no rest frame.

ohhh...

It is a fundamental principle of relativity that the speed of light has the same value c in every frame.

But if light had a rest frame the speed of light would have to be zero in that frame, otherwise it wouldn't be the light's rest frame.

So if light had a rest frame this would be a frame where the speed of light wasn't c, and that contradicts the fundamental principle on which SR is built.

6:43 AM

ohh

I dont have a clear view on SR though

is it simple or much complicated ?

The thing about SR is that the maths is really simple. It doesn't even require calculus.

But ...

The basic ideas are really, really hard to get your head around.

but?

@JohnRennie nice....

There isn't any easy way into the theory. What happens with students is that as they get more and more experience with the theory it gradually starts to make sense.

I can try and explain what I think is the most basic principle in SR if you want ...

@JohnRennie yes :D

OK. Go back to regular Newtonian mechanics. Suppose I have an arrow one metre long. I place it with one end on my origin and the other end at some point (x,y,z).

The end could be at any point (x,y,z) but whatever that point is the length of the arrow is given by Pythagoras' theorem so that means x² + y² + z² = 1.

OK so far?

6:49 AM

yes

just a min sir, I will reconnect to another netwok, the current one is very slow

Now, you measure the arrow in your coordinate system. Your coordinates could be translating or rotating relative to mine but for convenience we will assume we have the same origin i.e. the point (0,0,0) is the same in both our coordinates.

yes

So in your coordinates you have one end of the arrow at (0,0,0) like me, but the other end is at (x', y', z') where your x', y', z' are not the same as my x, y, z because we are using different coordinate systems.

Does this make sense so far?

yes

Cool :-) Now suppose you measure the length of the arrow using L = x'² + y'² + z'².

6:53 AM

@JohnRennie I guess this is where length contraction starts ?

You have to also get the answer 1 m because, well, it's the same arrow and in Newtonian mechanics lengths cannot change. Yes?

yes

@KavinIshwaran We are just using Newtonian mechanics so far, so lengths cannot change.

ohhh ok , clear

OK. The point of all this is that if we write L² = x² + y² + z² then the value of L is the same for all observers. It is what we call an invariant.

And it turns out this simple equation defines Euclidean geometry.

6:55 AM

yes

When Euclid was writing his books a few thousand years ago he could have started with the fact that x² + y² + z² is an invariant, and everything follows from there.

We normally write the equation as ds² = dx² + dy² + dz² where the dx etc mean an infinitesimal displacement.

And this equation is called the metric.

yes, sir can you wait for 5 mins, since I am encountering some network problems

OK

7:10 AM

@JohnRennie sir

shall we continue ?

Give me a moment ...

yes sir

oK

The point I'm working up to is that the metric is the equation that defines geometry. In this case the Euclidean metric ds² = dx² + dy² + dz² defines Euclidean geometry.

You may be wondering what the connection to SR is, but the geometry is the measurement of distances and angles so if we get things that change distances, like Lorentz contraction, this will change the geometry and hence the metric must be different.

So although it may not seem that way at first, relativity is actually a geometric theory.

yes....

But we'll get back to that ina moment.

Now, you probably know that in relativity we replace three dimensional space by four dimensional spacetime, so we have to treat time like a dimension. Yes?

7:16 AM

yes

That means our metric has to include time i.e. it has to compute the length of the vector (dt, dx, dy, dz) where dt is included. Yes?

yes

The obvious way to do this would be to write ds² = dt² + dx² + dy² + dz²

But in fact this is wrong.

Firstly there's a relatively trivial problem. In physics you cannot add quantities that have different units so you cannot add seconds to metres. That means the equation as written doesn't make sense.

yes

But suppose we multiply time by a constant that has the units of a velocity. Velocity × time is a distance. Yes?

7:20 AM

yes

So we could write ds² = c²dt² + dx² + dy² + dz²

where c is a constant with the units of a velocity.

Tis would now make sense. Yes?

yes.....

But, it turns out this is still wrong. The actual equation that describes our universe is:

ds² = -c²dt² + dx² + dy² + dz²

Note the minus sign in -c²dt²

where the minus comes from ?

That's a good question, and the answer is that we cannot derive this equation with the minus sign from first principles. When we do experiments we find that the equation must have a minus sign, and that's just the way the universe is.

The minus sign is there because ... well ... that's just the way the universe is.

7:24 AM

@JohnRennie so its empirical ?

Yes

Now, if you want I can show you how time dilation arises from this equation.

yes sir pls

OK. The key point is that ds is the length of the four vector (dt, dx, dy, dz) and just like in my example of an arrow this length must be same for all observers. The length cannot change just because we use a different coordinate system for measuring it.

yes

So if we consider two observers, and they both calculate ds² using their coordinates they must get the same value.

Now suppose I have a light that flashes every t seconds. I'm holding the light so in my frame it is at rest at the origin. In my coordinates the first flash is at (0,0,0,0) and the second flash is at (t,0,0,0). OK so far.

7:33 AM

yes

Now, you are moving along the x axis at a speed v relative to me. This means your x' coordinate will be different from mine. Our y and z coordinates will be the same so from now on I'm going to ignore them and just use t and x.

So in my frame the two flashes are at (0,0) and (t,0) and that means I get the length ds² to be:

ds² = -c²dt²

OK so far?

yes

We'll assume our origins coincide at time zero, so you also measure the first flash as happening at (0,0). Then you measure the second flash to happen at a point (T, X) where we have to work out what T and X are.

yes

In my frame you are moving at a speed v, so in your frame I am moving relative to you at a speed -v. Yes?

7:38 AM

yes

And that means in a time T I move a distance -vT, so the second flash happens at the point (T -vT) in your coordinates. So we can eliminate the variable X and we just have to figure out that T is.

Now you use the equation:

ds² = -c²dt² + dx²

And you substitute dt = T, dx = -vT, and you get:

ds² = -c²T² + v²T²

OK so far?

yes

But remember that ds² has to be the same for both of us. So we have:
ds² = -c²t² for me
ds² = -c²T² + v²T² for you

And if ds² is the same for both of us that must mean:

-c²t² = -c²T² + v²T²

Yes?

yes

And a quick rearrangement later we get:

T² = t²/(1 - v²/c²)

Yes?

7:44 AM

yes

But T is the time you measured between the flashes and t is the time I measured between the flashes, and we find the two times are not the same!

i.e. our clocks must be running at different rates. Yes?

definitely

And that's what time dilation is.

So just from the metric equation we predict time dilation.

Is this the way einstein too predicted it ?

No, in fact Einstein used a more complicated approach. It was a scientist called Minkowski who realised that special relativity could be written using the metric, and in fact the equation is called the Minkowski metric in his honour.

Minkowski space

In mathematical physics, Minkowski space (or Minkowski spacetime) () is a combination of three-dimensional Euclidean space and time into a four-dimensional manifold where the spacetime interval between any two events is independent of the inertial frame of reference in which they are recorded. Although initially developed by mathematician Hermann Minkowski for Maxwell's equations of electromagnetism, the mathematical structure of Minkowski spacetime was shown to be implied by the postulates of special relativity.Minkowski space is closely associated with Einstein's theories of special relativity...

7:50 AM

ohhh...

Thank you very much for your explanations sir ! :D

I doubt it has helped you get a real understanding of SR because as I said before it's really hard to get your head round this stuff.

If you study it at university you'll find it gradually starts to make sense as you learn more about it.

It really helped me understand !, Definitely It made sense.. I dont know I f get to learn SR GR and stuffs in university. I simply wanted to learn these things out of my interest :)

Once you know about the metric it isn't that big a step going to general relativity.

What happens in GR is that the metric changes, but the general principles remain the same.

I see

10:42 AM

@JohnRennie Hi :)

Hi :-)

If you dont mind, What is your qualification ?

physics.stackexchange.com/users/1325/john-rennie?tab=profile

what quantum chemistry is about ?

11:21 AM

@KavinIshwaran Quantum chemistry is using quantum mechanics to calculate how molecules react with each other.

nice

It was good fun, though back then computers were much slower than they are now and we could do only very simple calculations.

computers werent exist ?

Agreed, nowadays computers are even doing complex calculus problems

I used to work on an IBM mainframe.

IBM 308X

The IBM 308X was a line of mainframe computers, the first model of which, the Model 3081 Processor Complex, was introduced November 12, 1980. It consisted of a 3081 Processor Unit with supporting units. Later models in the series were the 3083 and the 3084. The 3083 was announced March 31 and the 3084 on September 3, both in 1982. The IBM 308X line introduced the System/370 Extended Architecture. All three 308X systems, which IBM had marketed as "System/370-Compatibles," were withdrawn August 4, 1987. == IBM 3081 == The initial 3081 offered, the 3081D, was a 5 MIPS machine. The next offering, the...

similar to commodore ?

11:28 AM

Like this.

I wonder how ppl work with these machines back then

what kind of calculation it will perform ?

I heard my father used to have a lot of floppy disks that have only 1.44 MB each

now a days a simple pdf file is exceeding 1mb

computers will not even have gui back then, the os should be something similar to CMD

12:27 PM

@JohnRennie Hi :-) Can you please help me with this question?

12:52 PM

@hansika 10^-3 coulombs ?

1:33 PM

@KavinIshwaran we have to find the x-coordinate, answer is 10 metres

@hansika I can see.. I asked if the specified unit for the charge is coulombs

@KavinIshwaran ohh..sorry..yess

3:08 PM

I keep getting the answer as 2.5 for the above question..

3:20 PM

here's what I did: angular velocity $w$ is a constant, $w = qB/m$. At any time $v = v_0 e^{-\frac{\alpha}{m}t}$. So now answer is simply $$\int_0^{\infty} v_o e^{\frac{-\alpha}{m}t} \sin(\frac{qB}{m}t) dt$$

which can be evaluated using integration by parts... seems like overkill for such a problem but I don't see why it should go wrong...

4:06 PM

@AshishAhuja me too

although theres a much simpler way to do thus

we can simply write the equations of motion in the x and y directions:

$$m\ddot{x}=q\dot{y}B-\alpha\dot{x}$$

intgerate both sides, t= 0 to say $t_{0}$. Now notice the antiderivative of the LHS is $m\dot{x}$, which is 0 at both t=0 and $t_{0}$. The Right side integral is simply $qyB-\alpha x$. (x=y=0 at t=0).

so we get $0=qyB-\alpha x$, which implies $y=\alpha x/qB$

now the equation in the y direction: $$m\ddot{y}=-q\dot{x}B-\alpha y$$

again, integrate both sides. $\dot{y}$ is 0 finally and $v0$ initally

so $mv_{0}= qBx+ \alpha y$. Substitute $y$ using $y=\alpha x/qB$ and then you can solve for x

$$x=\dfrac{qBm}{\alpha^{2}+ (qB)^{2}}$$

oops, missed a factor of $v_{0}$

yes

4:22 PM

yes ok

although its 1 so the calculation still comes out to be 2.5... :/

is the 4 in the question thats written as a subscript below C the missing factor of 4? :P

hah :D

maybe the answer could be wrong.. since two different methods are leading to the same answer

yes, definitely

anyway yeah I think your method was the one the question was looking for, the integral I got was way too messy to calculate.

yeah its doable tho, $e^{ax} sin(bx)$ is one of the standard ones. Plus that method allows you to find x(t) in general

you can do this by complex no.s btw, instead of doing it by parts

4:29 PM

ahh yes I think I have seen that using complex numbers before, I can't really remember how.

integral (e^(ax)sin(bx))=== imag part of (integral (e^(ax)e^(ibx))))

ahh ok yes, that is wayyy neater than by parts